cho 5,4g Al vào dung dịch CuSO4 thu được 34,2g Al2(SO4)3 và 19,2g Cu
a)lập PTHH b)tính mCuSO4 phản ứng c)tính số phân tử CuSO4
d)để có số phân tử HCl gấp 4 lần số phân tử CuSO4 cần lấy bao nhiêu gam HCl
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Ta có: \(n_{Ba\left(OH\right)_2}=0,4.0,5=0,2\left(mol\right)\)
PTHH: \(Ba\left(OH\right)_2+CuSO_4\rightarrow BaSO_4\downarrow+Cu\left(OH\right)_2\downarrow\)
0,2---------------------->0,2---------->0,2
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
0,2------------>0,2
`=>` \(m_{crắn}=0,2.80+0,2.233=62,6\left(g\right)\)
a) \(2Fe_3O_4+10H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}3Fe_2\left(SO_4\right)_3+SO_2\uparrow+10H_2O\)
b) Na2O (không phản ứng với NaOH nhưng phản ứng với H2O trong dd NaOH), Al2O3, P2O5 (không phản ứng với HCl nhưng phản ứng với H2O trong dd HCl)
\(Na_2O+H_2O\rightarrow2NaOH\\ Na_2O+2HCl\rightarrow2NaCl+H_2O\\ Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ P_2O_5+6NaOH\rightarrow2Na_3PO_4+3H_2O\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\)
c) SiO2
Ta có: \(\left\{{}\begin{matrix}n_{Ca\left(OH\right)_2}=0,2.1,75=0,35\left(mol\right)\\n_{CaCO_3}=\dfrac{25}{100}=0,25\left(mol\right)\end{matrix}\right.\)
Gọi \(n_{Fe}=a\left(mol\right)\left(ĐK:a>0\right)\)
Gọi CTHH của oxit sắt là \(Fe_xO_y\left(x,y\in N\text{*}\right)\)
PTHH:
\(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\) (1)
\(Fe_xO_y+yCO\xrightarrow[]{t^o}xFe+yCO_2\) (2)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\) (3)
\(2CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(HCO_3\right)_2\) (4)
\(Fe+H_2SO_{4\left(l\right)}\rightarrow FeSO_4+H_2\) (5)
a----------------------------------->a
Ta có: \(m_{ddH_2SO_4.tăng}=m_{Fe}-m_{H_2}\)
`=> 56a - 2a = 10,8 => a = 0,2 (mol)`
`-` TH1: Kết tủa chưa bị hòa tan
Theo PT (3): \(n_{Ca\left(OH\right)_2\left(pư\right)}=n_{CaCO_3}=0,25\left(mol\right)< 0,35\left(mol\right)\)
`=> Ca(OH)_2` dư (t/m)
Theo PT (3): \(n_{CO_2}=n_{CaCO_3}=0,25\left(mol\right)\)
Theo PT (1), (2): \(n_{O\left(oxit\right)}=n_{CO_2}=0,25\left(mol\right)\)
`=>` \(m_{Cu}=28-0,2.56-0,25.16=12,8\left(g\right)\)
`=>` \(n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)
`=>` \(n_{O\left(CuO\right)}=n_{CuO}=n_{Cu}=0,2\left(mol\right)\)
`=>` \(n_{O\left(Fe_xO_y\right)}=0,25-0,2=0,05\left(mol\right)\)
`=>` \(\dfrac{x}{y}=\dfrac{n_{Fe}}{n_O}=\dfrac{0,2}{0,05}=\dfrac{4}{1}\)
`=>` Loại
`-` TH2: Kết tủa đã bị hòa tan 1 phần
BTNT Ca: \(n_{Ca\left(HCO_3\right)_2}=n_{Ca\left(OH\right)_2}-n_{CaCO_3}=0,1\left(mol\right)\)
BTNT C: \(n_{CO_2}=2n_{Ca\left(HCO_3\right)_2}+n_{CaCO_3}=0,45\left(mol\right)\)
Theo PT (1), (2): \(n_{O\left(oxit\right)}=n_{CO_2}=0,45\left(mol\right)\)
`=>` \(m_{Cu}=28-0,45.16-0,2.56=9,6\left(g\right)\)
`=>` \(n_{O\left(CuO\right)}=n_{CuO}=n_{Cu}=\dfrac{9,6}{64}=0,15\left(mol\right)\)
`=>` \(n_{O\left(Fe_xO_y\right)}=0,45-0,15=0,3\left(mol\right)\)
`=>` \(\dfrac{x}{y}=\dfrac{n_{Fe}}{n_O}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\)
`=>` CTHH của oxit là Fe2O3
Gọi \(\left\{{}\begin{matrix}n_R=a\left(mol\right)\\n_{RO}=b\left(mol\right)\\n_{RCO_3}=c\left(mol\right)\end{matrix}\right.\left(ĐK:a,b,c>0\right)\)
`=>` \(aM_R+b\left(M_R+16\right)+c\left(M_R+60\right)=60,45\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{375.19,6\%}{98}=0,75\left(mol\right)\\n_{khí}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\end{matrix}\right.\)
PTHH: \(R+H_2SO_4\rightarrow RSO_4+H_2\) (1)
a----->a----------->a------>a
\(RO+H_2SO_4\rightarrow RSO_4+H_2O\) (2)
b------>b----------->b
\(RCO_3+H_2SO_4\rightarrow RSO_4+CO_2\uparrow+H_2O\)
c--------->c------------>c-------->c
`=>` \(\left\{{}\begin{matrix}\left(a+b+c\right)\left(M_R+96\right)=104,65\left(2\right)\\a+c=0,45\\n_{H_2SO_4\left(dư\right)}=0,75-a-b-c\left(mol\right)\end{matrix}\right.\)
Ta có: \(m_{ddspư}=375+60,45-2a-44c=435,45-2a-44c\left(g\right)\)
`=>` \(C\%_{H_2SO_4\left(dư\right)}=\dfrac{98\left(0,75-a-b-c\right)}{435,45-2a-44c}.100\%=2,311\%\left(4\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right),\left(4\right)\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\\c=0,25\\M_R=65\left(g/mol\right)\end{matrix}\right.\)
`=> A: Zn`
`=>` \(A:\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{ZnO}=0,2.81=16,2\left(g\right)\\m_{ZnCO_3}=0,25.125=31,25\left(g\right)\end{matrix}\right.\)
`-` TN1: Chọn \(m_{ddA}=m_{ddB}=100\left(g\right)\)
`=>` \(\left\{{}\begin{matrix}n_{HNO_3}=\dfrac{100.a\%}{63}=\dfrac{a}{63}\left(mol\right)\\n_{Ba\left(OH\right)_2}=\dfrac{100.b\%}{171}=\dfrac{b}{171}\left(mol\right)\end{matrix}\right.\)
Ta có: \(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\)
PTHH: \(Ba\left(OH\right)_2+2HNO_3\rightarrow Ba\left(NO_3\right)_2+2H_2O\)
\(\dfrac{a}{126}\)<------\(\dfrac{a}{63}\)
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
0,1<---------0,1
`=>` \(n_{Ba\left(OH\right)_2\left(dư\right)}=\dfrac{b}{171}-\dfrac{a}{126}=0,1\left(1\right)\)
`-` TN2: Chọn \(\left\{{}\begin{matrix}m_{ddA}=200\left(g\right)\\m_{ddB}=100\left(g\right)\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}n_{HNO_3}=\dfrac{200.a\%}{63}=\dfrac{2a}{63}\left(mol\right)\\n_{Ba\left(OH\right)_2}=\dfrac{100.b\%}{171}=\dfrac{b}{171}\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ba\left(OH\right)_2+2HNO_3\rightarrow Ba\left(NO_3\right)_2+2H_2O\)
\(\dfrac{b}{171}\)------>\(\dfrac{2b}{171}\)--------->\(\dfrac{b}{171}\)
`=>` \(\left\{{}\begin{matrix}m_{Ba\left(NO_3\right)_2}=\dfrac{261b}{171}=\dfrac{29b}{19}\left(g\right)\\m_{HNO_3\left(dư\right)}=63.\left(\dfrac{2a}{63}-\dfrac{2b}{171}\right)=2a-\dfrac{14b}{19}\left(g\right)\end{matrix}\right.\)
`=>` \(\dfrac{m_{Ba\left(NO_3\right)_2}}{m_{HNO_3}}=\dfrac{\dfrac{29b}{19}}{2a-\dfrac{14b}{19}}=\dfrac{87}{14}\)
`=>` \(\dfrac{406b}{19}=174a-\dfrac{1218b}{19}\)
`=>` \(174a-\dfrac{1624b}{19}=0\left(g\right)\)
Từ `(1), (2) =>` \(\left\{{}\begin{matrix}a=25,2\%\\b=51,3\%\end{matrix}\right.\)
A1: S
A2: Cu
A3: SO2
A4: NaHSO3
A5: Na2SO3
A6: BaSO3
A7: Ba(OH)2
(1) \(S+O_2\xrightarrow[]{t^o}SO_2\)
(2) \(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)
(3) \(SO_2+NaOH\rightarrow NaHSO_3\)
(4) \(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)
(5) \(Na_2SO_3+SO_2+H_2O\rightarrow2NaHSO_3\)
(6) \(Ba\left(OH\right)_2+2NaHSO_3\rightarrow BaSO_3\downarrow+Na_2SO_3+2H_2O\)
(7) \(Ba\left(OH\right)_2+Na_2SO_3\rightarrow BaSO_3\downarrow+2NaOH\)
a) PTHH: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
b) Theo ĐLBTKL: \(m_{Al}+m_{CuSO_4}=m_{Al_2\left(SO_4\right)_3}+m_{Cu}\)
`=>` \(m_{CuSO_4}=34,2+19,2-5,4=48\left(g\right)\)
`=>` \(n_{CuSO_4}=\dfrac{48}{160}=0,3\left(mol\right)\)
`=>` Số phân tử CuSO4: \(0,3.6.10^{23}=18.10^{22}\) (phân tử)
d) Số phân tử HCl = 4 số phân tử CuSO4
`=>` \(n_{HCl}=4n_{CuSO_4}=4.0,3=1,2\left(mol\right)\)
`=>` \(m_{HCl}=1,2.36,5=43,8\left(g\right)\)